On Parallel TransportDoes "parallel transport" move a vector parallel to itself on a curved surface even in the infinitesimal sense? You may think of two adjacent tangent planes on a curved surface.Is it always possible to have parallel vectors at the points of contact(one vector being preassigned) even if the planes are awkwardly inclined?

Dec4

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On Parallel TransportAt the point N'(referring to the original posting)you may consider a second vector tangent to the latitude-line at N'. If it(2nd vector) is parallel transported along the latitude to the point N" on the meridian NB, it is no more a tangent wrt to the latitude at N". The angle this vector makes with the tangent at N" should be equal to the angle which the first vector(moved up from the equator) makes with the meridian at N". This angle is expected to be small

On Parallel TransportPoints to Observe:(1)The vector after going through a loop[by parallel transport] rotates by a negligible small angle though the enclosed area is large. You will find this in the example I have referred to in the comment after the question after Lubos Moti's comment.(2)An infinitesimally small area is not sufficient to warrant a flat space-time.The Christoffel symbols are point functions. They have non-zero value for curved spacetime.

Dec3

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On Parallel TransportI am referring to the change between the initial and the final positions of the vector when it goes round a loop on a curved surface(by parallel transport). You may connect the point N'(in the original posting) with some point on the meridian NB by a $small{\;\;}$ curve so that the transported vector on landing on the meridian becomes tangential parallel or nearly tangential to the meridian NB.

Dec3

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On Parallel Transport(in continuation) The above idea is embodied in the non-zero value of the Christoffel tensors in curved space.

Dec3

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On Parallel TransportLet's's consider the vector components $A^\gamma(x^\alpha)$ and $ A^\gamma(x^\alpha+d x^\alpha)$. They have an infinitesimally small separation.This is not indicative of flat space-time over the infinitesimally small spacetime region concerned .Reason:For the purpose of calculating the derivative we have to parallel transport the vector-component $A(x^\alpha +d x^\alpha)$ to the location $x^\alpha$ and this vector definitely changes its orientation wrt to its initial position even though it has moved through an infinitesimally small distance

Dec2

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On Parallel TransportYou may consider a small curved line from N' to the meridian NB so that the vector becomes parallel to the meridian NB on reaching it. At A there is no turning all due to the exclusion of a small area.

Dec2

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On Parallel TransportQuoting Ron Maimon:"It ends up in nearly exactly the same direction as it was when you start the parallel transport"---in such a situation if you move back to A the vector does not turn by alpha. It turns by a much smaller amount! On removing the triangle the vector in the initial and the final situation (at A) make a very small angle. If the triangle is there it turns by 90 degrees after looping round

Dec2

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On Parallel TransportOnly if you consider a geodesic the an infinitesimally think space round it and parallel to it is nearly flat--the tangent vector propagates parallel to itself